powerpoint presentation

1. SUBTITLE

2.  Example 1. Determine the possible zeros or roots of the
polynomial function f(x) = 2x3 – x2 – 13x – 6.
Find the factors of -6 (represented by m) = {±1, ±2, ±3, ±6}
Find the factors of 2 (represented by (n) = {±1, ±2}
By using the Rational Theorem,
𝑀
𝑁
=
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 −6
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 2
=
{±1,±2,±3,±6}
{±1,±2}
To find the zeros or roots of a polynomial
function f(x), we use the Rational Theorem

3. 𝑀
𝑁
=
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑓𝑎𝑐𝑡𝑜𝑟𝑠 𝑜𝑓 −6
𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙𝑠 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 2
=
{±1,±2,±3,±6}
{±1,±2}
𝑀
𝑁
=
{±1,±2,±3,±6}
±1
,
{±1,±2,±3,±6}
±2
𝑀
𝑁
= { ± 1, ±2, ±3, ±6} , ±
1
2
, ±1, ±
3
2
, ±𝟑
𝑀
𝑁
= ±
1
2
, ±1, ±
3
2
, ±2, ±3, ±6 simplified
Therefore, the possible roots or zeros of f(x) = 2x3 – x2 – 13x – 6
is the set
𝑴
𝑵
= ±
𝟏
𝟐
, ±𝟏, ±
𝟑
𝟐
, ±𝟐, ±𝟑, ±𝟔 .

4. Steps Solution
1. Consider the possible zeros or
roots of f(x) = 2x3 – x3 – 13x – 6.
𝑀
𝑁
= ±
1
2
, ±1, ±
3
2
, ±2, ±3, ±6
2. By trial and error method,
determine the roots or zeros
using synthetic division
(depressed equation)
−𝟐 2 -1 -13 -6
-4 10 6
2 -5 -3 0
= (2x2 – 5x – 3)(quotient)
Example 2. Find the zeros or roots of f(x) = 2x3 – x2 – 13x – 6.

5. Steps Solution
3. Since -2 is one of the roots of
f(x) = 2x3 – x3 – 13x – 6,
we can write it as (x + 2).
Factor (2x2 – 5x – 3) using the
quadratic formula.
a= 2, b = -5, c = -3
x =
−𝑏± 𝑏2−4𝑎𝑐
2𝑎
x =
−(−5)± (−5)2−4(2)(−3)
2(2)
x =
5± 25+24
4
x =
5± 49
4
=
5±7
4
x =
5+7
4
=
12
4
= 3
x =
5 −7
4
=
−2
4
= -
𝟏
𝟐

6. Therefore, from the set
𝑀
𝑁
= ±
1
2
, ±1, ±
3
2
, ±2, ±3, ±6 of
possible zeros or roots of f(x) = 2x3 – x3 – 13x – 6, the actual
zeros or roots are -2, -
𝟏
𝟐
, and 3.

7. Solution
1. f(x) = 2x3 + 4x2 – 2x – 4
M = ±1, ±2, ±4
N = ±1, ±2
𝑀
𝑁
=
±1,±2,±4
±1
,
±1,±2,±4
±2
𝑀
𝑁
= ±1, ±
1
2
, ±2, ±4
Depressed equation:
(Using synthetic division)
𝟏 2 4 -2 -4
2 6 4
2 6 4 0
= (2x2 + 6x + 4)
Example 3. Find the zeros of a polynomial
f(x) = 2x3 + 4x2 – 2x – 4.

8. Solution
= (2x2 + 6x + 4)
Using factoring by groupings
in solving Quadratic Equation
a = 2, b = 6, c = 4
Multiply(a)(c) = (2)(4) = 8
Find the factors of 8 that gives
the sum of b (which is 6) = 4 . 2
= 2x2 + 6x + 4
= 2x2 + 4x + 2x + 4 = 0
(2x2 + 4x ) + (2x + 4) = 0
2x(x + 2) + 2(x + 2) = 0
(x + 2)(2x + 2) = 0
x + 2 = 0 and 2x + 2 = 0
x = -2 and 2x = 0 – 2
2x = -2
x = -1
Therefore the zeros of the
polynomial are 1, -1 and -2.